Question

A flat belt of negligible mass and thickness drives an output pulley of 0.6 m diameter, rotating at 500 rpm without slip. The arc of contact is \(190^{\circ}\) for this pulley, while it is \(170^{\circ}\) for the smaller input pulley. The coefficient of friction between the belt and the pulleys is 0.30. The measured slack side tension is 250 N. Neglecting other losses, the power transmitted by the belt drive, in kW, is nearest to (take \(\pi = 3.14\))

• A. 4.8
• B. 5.6
• C. 6.5
• D. 7.3

Hint:

Identify which pulley has the smaller angle of contact, since that one governs belt slip.

Answer 1

The Correct Option is B

Step 1: Use the common log form of the belt friction equation.
Textbooks often write $\dfrac{\mu\theta}{2.3026} = \log_{10}\left(\dfrac{T_1}{T_2}\right)$, which avoids working with $e^x$ directly.
The governing angle is the smaller wrap of $170^{\circ}$, since that pulley reaches the friction limit first.
$\theta = 170 \times \dfrac{3.14}{180} = 2.971$ rad, so $\mu\theta = 0.30 \times 2.971 = 0.8913$.

Step 2: Solve for the tension ratio.
$\log_{10}\left(\dfrac{T_1}{T_2}\right) = \dfrac{0.8913}{2.3026} = 0.3870$.
$\dfrac{T_1}{T_2} = 10^{0.3870} = 2.438$, so $T_1 = 2.438 \times 250 = 609.6$ N.

Step 3: Bring in the belt speed and net pulling force.
Belt speed $v = \dfrac{\pi D N}{60} = \dfrac{3.14 \times 0.6 \times 500}{60} = 15.70$ m/s.
Net driving force $= T_1 - T_2 = 609.6 - 250 = 359.6$ N.

Step 4: Compute the power.
$P = 359.6 \times 15.70 = 5646$ W, which is 5.65 kW.

Final Answer:
Rounded to the nearest listed value, the power transmitted is 5.6 kW, matching option B. \[ \boxed{P \approx 5.6\ \text{kW}} \]